# How to calculate the determinant bundle

Maybe, this is a problem of linear algebra. But I do not know how to calculate it. Let $E$ be a vector bundle of rank $2$ over an algebraic surface. If $H=S^{2n}E\bigotimes (\operatorname{det} E)^{\bigotimes -n}$, then $\operatorname{det} H$ is trivial, why?

## 2 Answers

Take a local basis $e_1,e_2$ for $E$, and then get a local basis $E_0=e_1^{2n}, E_1=e_1^{2n-1}e_2, \dots, E_{2n}=e_2^{2n}$ for $S^{2n}E$. If we scale $e_1$ by $a_1$ and $e_2$ by $a_2$, we scale $E_j$ by $a_1^{2n-j}a_2^j$. So we scale $E_0 \wedge \dots \wedge E_{2n}$ by $a_1^p a_2^p$ where $p=0+1+2+\dots+2n=(2n+1)n$. But in $S^{2n}E \otimes \det E^{\otimes -n}$, we replace each $E_j$ by something with an extra $a_1^{-n}a_2^{-n}$, say $E'_j = E_j \otimes (e_1 \wedge e_2)^{-\otimes n}$ scales by $a_1^{n-j} a_2^{j-n}$. So $E_0' \wedge \dots \wedge E_{2n}'$ scales by $a_1^p a_2^p$ where $p=n+(n-1)+\dots+(-n+1)+(-n)=0$. So $S^{2n}E \otimes \det E^{\otimes -n}$ has local section $E_0' \wedge \dots \wedge E_{2n}'$ invariant under these rescalings. It is easy to see that it is also invariant under adding multiples of $e_2$ to $e_1$ and vice versa, because of the wedge products, so that gives us invariance of our section under all linear transformations of $e_1$, $e_2$ with coefficients locally defined holomorphic functions. But then any two such sections must agree on overlaps, since the sections of $E$ will agree up to linear transformations.

This is a (very) easy application of the splitting principle. In fact, in order to compute the Chern classes one may assume $E=L_1 \oplus L_2$, where $L_1$ and $L_2$ are line bundles. Then by straightforward computations one has $$\det S^{2n}E = (\det E)^{\frac{2n(2n+1)}{2}} = (\det E)^{n(2n+1)}.$$ On the other hand, since $S^{2n}E$ has rank $2n+1$, for any line bundle $M$ one obtains by the same method $$\det (S^{2n} E \otimes M)= (\det S^{2n} E) \otimes M^{2n+1}= (\det E)^{ n(2n+1)} \otimes M^{2n+1}.$$ Taking $M= (\det E)^{-n}$ we are done.

• But, we know that line bundle is not determined by its Chern class. (unless on $\mathbb{P}^n$, you know for example) Sep 24 '13 at 11:07
• Of course. I should have said "in order to compute the Chern classes and the determinant of the symmetric powers of $E$", or just "in order to compute the determinant of the symmetric powers of $E$", since actually we are not interested in Chern classes. Sep 24 '13 at 12:00